Gravity sling. What is a gravity maneuver? How to fly in space and what is gravity

If a rocket flies near a planet, its speed will change. Either decrease or increase. It depends on which side of the planet it flies by.

When the US spacecraft Voyagers made their famous Grand Tour of the outer solar system, they performed several so-called gravity assist maneuvers near the giant planets.
The most fortunate was Voyager 2, which flew past all four major planets. The graph of its speed, see the figure:

The graph shows that after each approach to the planet (except for Neptune), the speed of the spacecraft increased by several kilometers per second.

At first glance, this may seem strange: an object flies into a gravitational field and accelerates, then flies out of the field and slows down. The arrival speed must be equal to the departure speed. Where does the extra energy come from?
Additional energy appears because there is a third body - the Sun. When flying near a planet, a spacecraft exchanges momentum and energy with it. If during such an exchange the gravitational energy of the planet in the field of the Sun decreases, then the kinetic energy of the spacecraft (SC) increases, and vice versa.

How should the spacecraft fly past the planet in order for its speed to increase? It is not difficult to answer this question. Let the spacecraft cross the orbit of the planet directly in front of it. In this case, having received an additional impulse in the direction of the planet, it will give it an additional impulse in the opposite direction, that is, in the direction of its movement. As a result, the planet will move to a slightly higher orbit, and its energy will increase. In this case, the energy of the spacecraft will correspondingly decrease. If the spacecraft crosses the orbit behind the planet, then it, having slightly slowed down its movement, will transfer the planet to a lower orbit. In this case, the speed of the spacecraft will increase.

Of course, the mass of the spacecraft is incommensurable with the mass of the planet. Therefore, the change in the orbital parameters of the planet during the gravitational maneuver is an infinitesimal value that cannot be measured. However, the energy of the planet is changing, and we can verify this by performing a gravity assist and seeing that the speed of the spacecraft changes. Here, for example, is how Voyager 2 flew near Jupiter on July 9, 1979 (see fig.). When approaching Jupiter, the speed of the spacecraft was 10 km/sec. At the moment of closest approach, it increased to 28 km/sec. And after Voyager 2 took off from the gravitational field of the gas giant, it decreased to 20 km / s. Thus, as a result of the gravitational maneuver, the speed of the spacecraft doubled and became hyperbolic. That is, it exceeded the speed necessary for departure from solar system. In the orbit of Jupiter, the speed of departure from the solar system is about 18 km / s.

This example shows that Jupiter (or another planet) can accelerate any body to hyperbolic speed. So, he can "throw" this body out of the solar system. Maybe modern cosmogonists are right? Maybe the giant planets really threw ice blocks to the far outskirts of the solar system and, thus, formed the cometary Oort cloud.
Before answering this question, let's see what kind of gravitational maneuvers the planets are capable of?

2. Principles of gravity assist

I first got acquainted with the gravitational maneuver in the 9th grade at the regional Olympiad in physics. The task was this. A rocket is launched from the earth at a speedV(enough to fly out of the gravity field). The rocket has an engine with thrust F which can work time t. At what point in time must the engine be turned on so that the final speed of the rocket is maximum? Ignore air resistance.

At first it seemed to me that it did not matter when to turn on the engine. After all, due to the law of conservation of energy, the final speed of the rocket must be the same in any case. It remained to calculate the final speed of the rocket in two cases: 1. we turn on the engine at the beginning, 2. we turn on the engine after leaving the Earth's gravity field. Then compare the results and make sure that the final speed of the rocket is the same in both cases. But then I remembered that power is equal to: traction force times speed. Therefore, the power of the rocket engine will be maximum if the engine is turned on immediately at the start, when the rocket speed is maximum. So, the correct answer is: we turn on the engine immediately, then the final speed of the rocket will be maximum.

And although I solved the problem correctly, but the problem remained. The final speed, and, therefore, the energy of the rocket DEPENDS on at what point in time the engine is turned on. It seems to be a clear violation of the law of conservation of energy. Or not? What's the matter here? Energy must be conserved! I tried to answer all these questions after the Olympiad.

May we have a mass rocket M with an engine that creates thrust by force F. Let's place this rocket in empty space (away from stars and planets) and turn on the engine. How fast will the rocket move? We know the answer from Newton's second law: acceleration a equals:

a=F/M

The mass of the rocket also does not change (100 km / s is not yet a relativistic case), so the thrust force F will be the SAME. And, therefore, the power of the rocket DEPENDS on its speed. After all, power equals force times speed. It turns out that if a rocket is moving at a speed of 100 km / s, then the power of its engine is 100 times more powerful than EXACTLY THE SAME engine located on a rocket moving at a speed of 1 km / s.

At first glance, this may seem strange and even paradoxical. Where does the huge extra power come from? Energy must be conserved!

Let's look into this issue.

A rocket always moves on jet thrust: it throws various gases into space at high speed. For definiteness, we assume that the speed of the emission of gases is 10 km/sec. If a rocket is moving at a speed of 1 km/sec, then its engine accelerates mainly not the rocket, but the propellant. Therefore, the engine power to accelerate the rocket is not high. But if the rocket is moving at a speed of 10 km / s, then the ejected fuel will be at rest relative to the external observer, that is, the entire engine power will be spent on rocket acceleration. And if the rocket is moving at a speed of 100 km / s? In this case, the ejected fuel will move at a speed of 90 km/sec. That is, the speed of the fuel WILL DECREASE from 100 to 90 km/s. And the ALL difference in the kinetic energy of the fuel, due to the law of conservation of energy, will be transferred to the rocket. Therefore, the power of the rocket engine at such speeds will increase significantly.

Simply put, a fast-moving rocket has a lot of kinetic energy in its propellant. And from this energy, additional power is drawn to accelerate the rocket. Now it remains to figure out how this property of the rocket can be used in practice.

3. Practical application

Suppose in the near future you are going to fly a rocket to the Saturn system to Titan:

to study anaerobic life forms.

They flew to the orbit of Jupiter and it turned out that the speed of the rocket had dropped to almost zero. They did not calculate the flight path properly or the fuel turned out to be counterfeit. Or maybe a meteorite hit the fuel bay, and almost all the fuel was lost. What to do?

“We enter the field of attraction of Jupiter and fall on it. As a result, Jupiter accelerates the rocket to a huge speed - about 60 km / s. When the rocket accelerates to this speed, turn on the engine. Engine power at this speed will increase many times. Then we take off from the field of attraction of Jupiter. As a result of such a gravitational maneuver, the speed of the rocket increases not by 1 km / s, but much more. And we can fly to Saturn."

But someone objects.

“Yes, the power of the rocket near Jupiter will increase. The rocket will receive additional energy. But, flying out of Jupiter's field of attraction, we will lose all this additional energy. The energy must remain in the potential well of Jupiter, otherwise there will be something like a perpetual motion machine, and this is impossible. Therefore, there will be no benefit from the gravitational maneuver. We're just wasting our time."

What do you think of it?

So, the rocket is not far from Jupiter and is almost motionless relative to it. The rocket has an engine with enough fuel to increase the rocket's speed by only 1 km/sec. To increase the efficiency of the engine, it is proposed to perform a gravitational maneuver: "drop" the rocket on Jupiter. She will move in his field of attraction along a parabola (see photo). And at the lowest point of the trajectory (marked with a red cross in the photo), turn on the engine. The speed of the rocket near Jupiter will be 60 km/sec. After the engine further accelerates it, the speed of the rocket will increase to 61 km / s. What speed will the rocket have when it leaves Jupiter's field of gravity?

This task is within the power of a high school student, if, of course, he knows physics well. First you need to write a formula for the sum of potential and kinetic energies. Then remember the formula for the potential energy in the gravitational field of the ball. Look in the reference book, what is the gravitational constant, as well as the mass of Jupiter and its radius. Using the law of conservation of energy and performing algebraic transformations, obtain a general final formula. And finally, substituting all the numbers into the formula and doing the calculations, get the answer. I understand that no one (almost no one) wants to delve into some formulas, so I will try, without straining you with any equations, to explain the solution of this problem “on the fingers”. Hope it works!

At the point marked with a cross:

the rocket speed is 60 km/sec, and the energy is 3600 units. 3600 units is enough to fly out of Jupiter's field of attraction. After the rocket accelerated, its speed became 61 km / s, and the energy, respectively, 61 squared (we take the calculator) 3721 units. When a rocket flies out of Jupiter's field of gravity, it only consumes 3600 units. There are 121 units left. This corresponds to a speed (take the square root) of 11 km/sec. Problem solved. This is not an approximation, but an EXACT answer.

We see that the gravitational maneuver can be used to obtain additional energy. Instead of accelerating the rocket to 1 km / s, it can be accelerated to 11 km / s (121 times more energy, efficiency - 12 thousand percent!), If there is some massive body like Jupiter nearby.

Due to what we received a HUGE energy gain? Due to the fact that they left the spent fuel not in empty space near the rocket, but in a deep potential well created by Jupiter. The spent fuel received a large potential energy with a MINUS sign. Therefore, the rocket received a large kinetic energy with a PLUS sign.

4. Rotation of the velocity vector near the planet

In the very general view this task looks like this. We fly into Jupiter's gravitational field at some speed. Then we fly out of the field. Will our speed change? And how much can it change? Let's solve this problem.

From the point of view of an observer who is on Jupiter (or rather, stationary relative to its center of mass), our maneuver looks like this. At first, the rocket is at a great distance from Jupiter and moves towards it at a speed V. Then, approaching Jupiter, it accelerates. In this case, the rocket trajectory is curved and, as is known, in its most general form is a hyperbole. The maximum speed of the rocket will be at the minimum approach. The main thing here is not to crash into Jupiter, but to fly next to it. After the minimum approach, the rocket will begin to move away from Jupiter, and its speed will decrease. Finally, the rocket will fly out of Jupiter's field of gravity. What will her speed be? Exactly the same as it was on arrival. The rocket flew into the gravitational field of Jupiter with a speed V and flew out of it at exactly the same speed V. Nothing changed? No has changed. The DIRECTION of speed has changed. It is important. Thanks to this, we can perform a gravitational maneuver.

Indeed, what is important for us is not the speed of the rocket relative to Jupiter, but its speed relative to the Sun. This is the so-called heliocentric velocity. With such a speed, the rocket moves through the solar system. Jupiter also moves around the solar system. The rocket's heliocentric velocity vector can be decomposed into the sum of two vectors: Jupiter's orbital velocity (about 13 km/sec) and the rocket's velocity RELATIVE to Jupiter. There is nothing complicated here! This is the usual triangle rule for vector addition, which is taught in 7th grade. And this rule is ENOUGH to understand the essence of the gravity maneuver.

We have four speeds. V 1 is the speed of our rocket relative to the Sun BEFORE the gravity assist. U 1 is the speed of the rocket relative to Jupiter BEFORE the gravity assist. U 2 is the speed of the rocket relative to Jupiter AFTER the gravity assist. By size U 1 and U 2 are EQUAL, but in direction they are DIFFERENT. V 2 is the speed of the rocket relative to the Sun AFTER the gravity assist. To see how all these four speeds are related, let's look at the figure:

The green arrow AO is the speed of Jupiter in its orbit. The red arrow AB is V 1: The speed of our rocket relative to the Sun BEFORE the gravity assist. The yellow arrow OB is the speed of our rocket relative to Jupiter BEFORE the gravitational maneuver. The yellow OS arrow is the speed of the rocket relative to Jupiter AFTER the gravity assist. This speed MUST lie somewhere on the yellow circle of OB radius. Because in its coordinate system, Jupiter CANNOT change the value of the rocket's speed, but can only rotate it by a certain angle (alpha). And finally, AC is what we need: rocket speed V 2 AFTER the gravity assist.

See how simple it is. The speed of the rocket AFTER the gravity assist AC is equal to the speed of the rocket BEFORE the gravity assist AB plus the vector BC. And the BC vector is a CHANGE in the speed of the rocket in Jupiter's frame of reference. Because OS - OB = OS + IN = IN + OS = BC. The more the rocket's velocity vector rotates relative to Jupiter, the more effective the gravitational maneuver will be.

So, a rocket WITHOUT fuel flies into the gravitational field of Jupiter (or another planet). The magnitude of its speed BEFORE and AFTER the maneuver relative to Jupiter DOES NOT CHANGE. But due to the rotation of the velocity vector relative to Jupiter, the rocket's velocity relative to Jupiter still changes. And the vector of this change is simply added to the velocity vector of the rocket BEFORE the maneuver. I hope I explained everything clearly.

There is another way to accelerate an object to a speed close to the speed of light - to use the "sling effect". When sending space probes to other planets, NASA sometimes makes them maneuver around a neighboring planet in order to use the "sling effect" to further disperse the device. This is how NASA saves valuable rocket fuel. This is how the Voyager 2 spacecraft managed to fly to Neptune, whose orbit lies at the very edge of the solar system.

Freeman Dyson, a physicist at Princeton, made an interesting suggestion. If someday in the distant future, humanity will be able to detect in space two neutron stars, revolving around a common center at high speed, then an earthly ship, flying very close to one of these stars, can, due to a gravitational maneuver, gain speed equal to almost a third of the speed of light. As a result, the ship would accelerate to near-light speeds due to gravity. Theoretically, this could happen.

Only in reality this way of accelerating with the help of gravity will not work. (The law of conservation of energy says that a roller coaster cart, accelerating on the descent and slowing on the ascent, ends up at the top at exactly the same speed as at the very beginning - there is no increase in energy. Similarly, wrapping around the stationary Sun , we will finish at exactly the same speed as we started the maneuver.) The Dyson method with two neutron stars could in principle work, but only because neutron stars move fast. A spacecraft using a gravitational maneuver receives an increase in energy due to the movement of a planet or star. If they are motionless, such a maneuver will not work.

And Dyson's suggestion, while it might work, won't help today's scientists on Earth, because visiting fast-rotating neutron stars would first require building a starship.

From the gun to the sky

Another ingenious way to launch a ship into space and accelerate it to fantastic speeds is to shoot it from a rail electromagnetic “gun”, which Arthur C. Clarke and other science fiction authors described in their works. This project is currently being seriously considered as a possible part of the Star Wars missile shield.

The method consists in using the energy of electromagnetism to accelerate the rocket to high speeds instead of rocket fuel or gunpowder.

At its simplest, a rail gun is two parallel wires or rails; the rocket projectile, or missile, "sits" on both rails, forming a U-shaped configuration. Even Michael Faraday knew that a force acts on a frame with an electric current in a magnetic field. (Generally speaking, all electric motors work on this principle.) If an electric current of millions of amperes is passed through the rails and the projectile, an extremely powerful magnetic field will arise around the entire system, which, in turn, will drive the projectile along the rails, accelerate it to tremendous speed and throw it into space from the end of the rail system.

During tests, rail-mounted electromagnetic guns successfully fired metal objects at tremendous speeds, accelerating them over a very short distance. Remarkably, in theory, an ordinary rail gun is capable of firing a metal projectile at a speed of 8 km / s; this is enough to put it into low Earth orbit. In principle, the entire NASA rocket fleet could be replaced by rail guns, which would fire a payload into orbit directly from the surface of the Earth.

The railgun has significant advantages over chemical guns and rockets. When you fire a gun, the maximum speed at which the expanding gases can push the bullet out of the barrel is limited by the speed of the shock wave. Jules Berne in the classic novel "From the Earth to the Moon" shot a projectile with astronauts to the Moon using gunpowder, but in fact it is easy to calculate that the maximum speed that a powder charge can give a projectile is many times less than the speed needed to fly to the Moon . The railgun, on the other hand, does not use the explosive expansion of gases and therefore does not depend in any way on the speed of propagation of the shock wave.

But the railgun has its own problems. Objects on it are accelerating so fast that they tend to be flattened due to collision... with air. The payload is severely deformed when the railgun is fired from the muzzle, because when the projectile hits the air, it's like hitting a brick wall. In addition, during acceleration, the projectile experiences tremendous acceleration, which in itself is capable of greatly deforming the load. The rails must be replaced regularly, as the projectile also deforms them when moving. Moreover, overloads in a rail gun are fatal to humans; human bones simply can not withstand such acceleration and collapse.

One solution is to put a railgun on the moon. There, outside earth's atmosphere, the projectile will be able to accelerate unhindered in the vacuum of outer space. But even on the Moon, the projectile during acceleration will experience enormous overloads that can damage and deform the payload. In a sense, a railgun is the antithesis of a laser sail, which picks up speed gradually over time. The limitations of the rail gun are determined precisely by the fact that it transfers enormous energy to the body at a short distance and in a short time.

A railgun capable of firing a craft at the nearest stars would be a very expensive construction. Thus, one of the projects involves the construction of open space a rail gun two-thirds the distance from the earth to the sun. This gun would have to store solar energy and then use it all at once, accelerating a ten-ton payload to a speed equal to a third of the speed of light. In this case, the "projectile" will experience an overload of 5000 g. Of course, only the most enduring robot ships will be able to “survive” such a launch.

conventional view

There are special bodies in the solar system - comets.
A comet is a small body several kilometers in size. Unlike an ordinary asteroid, a comet contains various ices: water, carbon dioxide, methane and others. When the comet enters the orbit of Jupiter, these ices begin to evaporate rapidly, leave the surface of the comet together with dust and form the so-called coma - a gas and dust cloud surrounding the solid core. This cloud extends hundreds of thousands of kilometers from the core. Thanks to the reflected sunlight, the comet (not itself, but only a cloud) becomes visible. And due to light pressure, part of the cloud is pulled into the so-called tail, which stretches from the comet for many millions of kilometers (see photo 2). Due to very weak gravity, all the substance of the coma and tail is irretrievably lost. Therefore, flying near the Sun, a comet can lose several percent of its mass, and sometimes more. The time of her life by astronomical standards is negligible.
Where do new comets come from?

According to traditional cosmogony, they come from the so-called Oort cloud. It is generally accepted that at a distance of one hundred thousand astronomical units from the Sun (half the distance to the nearest star) there is a huge reservoir of comets. The nearest stars periodically disturb this reservoir, and then the orbits of some comets change so that their perihelion is near the Sun, the gases on its surface begin to evaporate, forming a huge coma and tail, and the comet becomes visible through a telescope, and sometimes even with the naked eye. Pictured is the famous Great Comet Hale-Bopp, in 1997.

How did the Oort cloud form? The generally accepted answer is this. At the very beginning of the formation of the solar system in the region of the giant planets, many icy bodies with a diameter of ten or more kilometers formed. Some of them became part of the giant planets and their satellites, and some were ejected to the periphery of the solar system. Jupiter played the main role in this process, but Saturn, Uranus and Neptune also applied their gravitational fields to it. In the most general terms, this process looked like this: a comet flies near the powerful gravitational field of Jupiter, and it changes its speed so that it ends up on the periphery of the solar system.

True, this is not enough. If the comet's perihelion is inside the orbit of Jupiter, and the aphelion is somewhere on the periphery, then its period, as it is easy to calculate, will be several million years. During the existence of the solar system, such a comet will have time to approach the Sun almost a thousand times and all of its gas that can evaporate will evaporate. Therefore, it is assumed that when the comet is on the periphery, then the perturbations from the nearest stars will change its orbit so that the perihelion will also be very far from the Sun.

So there is a four step process. 1. Jupiter throws a piece of ice to the periphery of the solar system. 2. The nearest star changes its orbit so that the perihelion of the orbit is also far from the Sun. 3. In such an orbit, a piece of ice stays safe and sound for almost several billion years. 4. Another passing star again perturbs its orbit so that the perihelion is near the Sun. As a result, a piece of ice flies towards us. And we see it like a new comet.

All this seems quite plausible to modern cosmogonists. But is it? Let's take a closer look at all four steps.

GRAVITY MANEUVER

First meeting

I first got acquainted with the gravitational maneuver in the 9th grade at the regional Olympiad in physics. The task was this.
A rocket is launched from the Earth at a speed V (sufficient to fly out of the field of gravity). The rocket has an engine with thrust F, which can operate for a time t. At what point in time must the engine be turned on so that the final speed of the rocket is maximum? Ignore air resistance.

Rocket thrust DEPENDS on its speed. This is an important point and worth discussing.
Suppose we have a rocket of mass M with an engine that creates thrust with force F. Let's place this rocket in empty space (away from stars and planets) and turn on the engine. How fast will the rocket move? We know the answer from Newton's Second Law: the acceleration A is equal to:
A = F/M

Now let's move on to another inertial frame of reference, in which the rocket moves at a high speed, say, 100 km/sec. What is the acceleration of the rocket in this frame of reference?
Acceleration DOES NOT DEPEND on the choice of inertial frame of reference, so it will be the SAME:
A = F/M
The mass of the rocket also does not change (100 km / s is not yet a relativistic case), so the thrust force F will be the SAME.
And, therefore, the power of the rocket DEPENDS on its speed. After all, power equals force times speed. It turns out that if a rocket is moving at a speed of 100 km / s, then the power of its engine is 100 times more powerful than EXACTLY THE SAME engine located on a rocket moving at a speed of 1 km / s.

At first glance, this may seem strange and even paradoxical. Where does the huge extra power come from? Energy must be conserved!
Let's look into this issue.
A rocket always moves on jet thrust: it throws various gases into space at high speed. For definiteness, we assume that the speed of the emission of gases is 10 km/sec. If a rocket is moving at a speed of 1 km/sec, then its engine accelerates mainly not the rocket, but the propellant. Therefore, the engine power to accelerate the rocket is not high. But if the rocket is moving at a speed of 10 km / s, then the ejected fuel will be at rest relative to the external observer, that is, the entire engine power will be spent on rocket acceleration. And if the rocket is moving at a speed of 100 km / s? In this case, the ejected fuel will move at a speed of 90 km/sec. That is, the speed of the fuel WILL DECREASE from 100 to 90 km/s. And the ALL difference in the kinetic energy of the fuel, due to the law of conservation of energy, will be transferred to the rocket. Therefore, the power of the rocket engine at such speeds will increase significantly.

Simply put, a fast-moving rocket has a lot of kinetic energy in its propellant. And from this energy, additional power is drawn to accelerate the rocket.

Now it remains to figure out how this property of the rocket can be used in practice.

An attempt at practical application

Suppose, in the near future, you are going to fly a rocket to the Saturn system on Titan (see photos 1-3) to study anaerobic life forms. They flew to the orbit of Jupiter and it turned out that the speed of the rocket had dropped to almost zero. The flight path was not calculated properly or the fuel turned out to be counterfeit :) . Or maybe a meteorite hit the fuel bay, and almost all the fuel was lost. What to do?

The rocket has an engine and a small amount of fuel left. But the maximum that the engine is capable of is to increase the speed of the rocket by 1 km / s. This is clearly not enough to fly to Saturn. And now the pilot offers such an option.
“We enter the field of attraction of Jupiter and fall on it. As a result, Jupiter accelerates the rocket to a tremendous speed - about 60 km / s. When the rocket accelerates to this speed, turn on the engine. Engine power at this speed will increase many times. Then we take off from the field of attraction of Jupiter. As a result of such a gravitational maneuver, the speed of the rocket increases not by 1 km / s, but much more. And we can fly to Saturn."
But someone objects.
“Yes, the power of the rocket near Jupiter will increase. The rocket will receive additional energy. But, flying out of Jupiter's field of attraction, we will lose all this additional energy. The energy must remain in the potential well of Jupiter, otherwise there will be something like a perpetual motion machine, and this is impossible. Therefore, there will be no benefit from the gravitational maneuver. We're just wasting our time."

So, the rocket is not far from Jupiter and is almost motionless relative to it. The rocket has an engine with enough fuel to increase the rocket's speed by only 1 km/sec. To increase the efficiency of the engine, it is proposed to perform a gravitational maneuver: "drop" the rocket on Jupiter. She will move in his field of attraction along a parabola (see photo). And at the lowest point of the trajectory (marked with a red cross in the photo) will turn on l engine. The speed of the rocket near Jupiter will be 60 km/sec. After the engine further accelerates it, the speed of the rocket will increase to 61 km / s. What speed will the rocket have when it leaves Jupiter's field of gravity?

If the rocket is stationary, its kinetic energy is zero. And if the rocket moves at a speed of 1 km / s, then we will assume that its energy is 1 unit. Accordingly, if the rocket moves at a speed of 2 km / s, then its energy is 4 units, if 10 km / s, then 100 units, etc. This is clear. We have already solved half of the problem.
At the point marked with a cross (see photo), the speed of the rocket is 60 km / s, and the energy is 3600 units. 3600 units is enough to fly out of Jupiter's field of attraction. After the rocket accelerated, its speed became 61 km / s, and the energy, respectively, 61 squared (we take the calculator) 3721 units. When a rocket flies out of Jupiter's field of gravity, it only consumes 3600 units. There are 121 units left. This corresponds to a speed (take the square root) of 11 km/sec. Problem solved. This is not an approximation, but an EXACT answer.

Vector rotation

Suppose we are flying a rocket near Jupiter and we want to increase its speed. But we don't have fuel. Let's just say we have some fuel to correct our course. But it is clearly not enough to noticeably disperse the rocket. Can we noticeably increase the speed of a rocket using gravity assist?
In its most general form, this task looks like this. We fly into Jupiter's gravitational field at some speed. Then we fly out of the field. Will our speed change? And how much can it change?
Let's solve this problem.

From the point of view of an observer who is on Jupiter (or rather, stationary relative to its center of mass), our maneuver looks like this. First, the rocket is at a great distance from Jupiter and moves towards it at a speed V. Then, approaching Jupiter, it accelerates. In this case, the rocket trajectory is curved and, as is known, in its most general form is a hyperbole. The maximum speed of the rocket will be at the minimum approach. The main thing here is not to crash into Jupiter, but to fly next to it. After the minimum approach, the rocket will begin to move away from Jupiter, and its speed will decrease. Finally, the rocket will fly out of Jupiter's field of gravity. What will her speed be? Exactly the same as it was on arrival. The rocket flew into the gravitational field of Jupiter at a speed V and flew out of it at exactly the same speed V. Has anything changed? No has changed. The DIRECTION of speed has changed. It is important. Thanks to this, we can perform a gravitational maneuver.

We have four speeds. U(1) is the speed of our rocket relative to the Sun BEFORE the gravity assist. V(1) is the speed of the rocket relative to Jupiter BEFORE the gravity assist. V(2) is the speed of the rocket relative to Jupiter AFTER the gravity assist. V(1) and V(2) are EQUAL in magnitude, but they are DIFFERENT in direction. U(2) is the speed of the rocket relative to the Sun AFTER the gravity assist. To see how all these four speeds are related, look at the figure.

The green arrow AO is the speed of Jupiter in its orbit. The red arrow AB is U(1): the speed of our rocket relative to the Sun BEFORE the gravity assist. The yellow arrow OB is the speed of our rocket relative to Jupiter BEFORE the gravitational maneuver. The yellow OS arrow is the speed of the rocket relative to Jupiter AFTER the gravity assist. This speed MUST lie somewhere on the yellow circle of OB radius. Because in its coordinate system, Jupiter CANNOT change the value of the rocket's speed, but can only rotate it by a certain angle (alpha). And finally, AC is what we need: U(2) rocket speed AFTER the gravity assist.

To better understand the essence of the gravitational maneuver, we will analyze it using the example of Voyager 2, which flew near Jupiter on July 9, 1979. As can be seen from the graph (see photo), he flew up to Jupiter at a speed of 10 km / s, and flew out of his gravitational field at a speed of 20 km / s. Only two numbers: 10 and 20.
You will be surprised how much information can be extracted from these numbers:
1. We will calculate what speed Voyager 2 had when it left the Earth's gravitational field.
2. Let's find the angle at which the apparatus approached Jupiter's orbit.
3. Calculate the minimum distance that Voyager 2 flew to Jupiter.
4. Let's find out what its trajectory looked like relative to an observer located on Jupiter.
5. Find the angle by which the spacecraft deviated after the encounter with Jupiter.

We will not use complex formulas, but will do the calculations, as usual, “on the fingers”, sometimes using simple drawings. However, the answers we get will be accurate. Let's just say they might not be exact, because the numbers 10 and 20 are most likely not exact. They are taken from the chart and rounded. In addition, other numbers that we will use will also be rounded. After all, it is important for us to understand the gravitational maneuver. Therefore, we will take the numbers 10 and 20 as exact, so that there is something to build on.

Let's solve the 1st problem.
Let's agree that the energy of Voyager-2 moving at a speed of 1 km/sec is 1 unit. The minimum departure speed from the solar system from the orbit of Jupiter is 18 km/sec. The graph of this speed is in the photo, but it is located like this. It is necessary to multiply the orbital speed of Jupiter (about 13 km / s) by the root of two. If Voyager 2, when approaching Jupiter, had a speed of 18 km / s (energy 324 units), then its total energy (the sum of kinetic and potential) in the gravitational field of the Sun would be EXACTLY equal to zero. But the speed of Voyager 2 was only 10 km / s, and the energy was 100 units. That is, less than:
324-100 = 224 units.
This lack of energy is CONTAINED as Voyager 2 travels from Earth to Jupiter.
The minimum departure speed from the solar system from the Earth's orbit is approximately 42 km / s (slightly more). To find it, you need to multiply the orbital speed of the Earth (about 30 km / s) by the root of two. If Voyager 2 were moving away from Earth at a speed of 42 km/sec, its kinetic energy would be 1764 units (42 squared) and the total would be ZERO. As we have already found out, the energy of Voyager 2 was less than 224 units, that is, 1764 - 224 = 1540 units. We take the root of this number and find the speed with which Voyager 2 flew out of the Earth's gravity field: 39.3 km / s.

When a spacecraft is launched from the Earth into the outer part of the solar system, it is launched, as a rule, along the orbital velocity of the Earth. In this case, the speed of the Earth's movement is ADDED to the speed of the apparatus, which leads to a huge gain in energy.

And how is the issue with the DIRECTION of speed solved? Very simple. They wait until the Earth reaches the desired part of its orbit so that the direction of its speed is the one that is needed. Say, when launching a rocket to Mars, there is a small “window” in time in which it is very convenient to launch. If, for some reason, the launch failed, then the next attempt, you can be sure, will not be earlier than two years later.

When at the end of the 70s of the last century the giant planets lined up in a certain order, many scientists - specialists in celestial mechanics suggested taking advantage of a happy accident in the location of these planets. A project was proposed on how to carry out the Grand Tour at minimal cost - a trip to ALL the giant planets at once. Which was done with success.
If we had unlimited resources and fuel, we could fly wherever we want, whenever we want. But since energy has to be saved, scientists carry out only energy-efficient flights. You can be sure that Voyager 2 was launched along the direction of the Earth's motion.
As we calculated earlier, its speed relative to the Sun was 39.3 km/sec. When Voyager 2 flew to Jupiter, its speed dropped to 10 km / s. Where was she sent?
The projection of this velocity onto Jupiter's orbital velocity can be found from the law of conservation of angular momentum. The radius of Jupiter's orbit is 5.2 times that of the Earth's orbit. So, you need to divide 39.3 km / s by 5.2. We get 7.5 km / s. That is, the cosine of the angle we need is 7.5 km / s (Voyager velocity projection) divided by 10 km / s (Voyager velocity), we get 0.75. The angle itself is 41 degrees. At this angle, Voyager 2 flew into the orbit of Jupiter.

Knowing the speed of Voyager 2 and the direction of its movement, we can draw a geometric diagram of the gravity assist. It is done like this. We select point A and draw from it the vector of Jupiter's orbital velocity (13 km / s on the selected scale). The end of this vector (green arrow) is denoted by the letter O (see photo 1). Then from point A we draw the velocity vector of Voyager 2 (10 km / s on the selected scale) at an angle of 41 degrees. The end of this vector (red arrow) is denoted by the letter B.
Now we build a circle (yellow color) with a center at point O and radius |OB| (see photo 2). The end of the velocity vector both before and after the gravitational maneuver can lie only on this circle. Now we draw a circle with a radius of 20 km/sec (in the chosen scale) centered at point A. This is Voyager's speed after the gravity assist. It intersects with the yellow circle at some point C.

We have drawn the gravity assist that Voyager 2 performed on July 9, 1979. AO is Jupiter's orbital velocity vector. AB is the velocity vector at which Voyager 2 approached Jupiter. Angle OAB is 41 degrees. AC is the velocity vector of Voyager 2 AFTER the gravity assist. It can be seen from the drawing that the angle OAC is approximately 20 degrees (half the angle OAB). If desired, this angle can be calculated exactly, since all the triangles in the drawing are given.
OB is the velocity vector at which Voyager 2 was approaching Jupiter, FROM THE POINT OF VIEW of an observer on Jupiter. OS - Voyager's velocity vector after the maneuver relative to the observer on Jupiter.

If Jupiter were not rotating and you were in the subsolar side (the Sun is at its zenith), then you would see Voyager 2 moving from West to East. First, it appeared in the western part of the sky, then, approaching, reached the Zenith, flying near the Sun, and then disappeared behind the horizon in the East. Its velocity vector has turned, as can be seen from the drawing, by about 90 degrees (angle alpha).

Gravity maneuver for accelerating an object Gravity maneuver for decelerating an object Gravity maneuver for accelerating, decelerating or changing the direction of a spacecraft flight, under the influence of gravitational fields celestial bodies.… … Wikipedia

Gravity maneuver to accelerate an object Gravity maneuver to slow down an object Gravity maneuver to accelerate, decelerate or change the direction of the flight of a spacecraft, under the influence of gravitational fields of celestial bodies ... ... Wikipedia

- ... Wikipedia

This is one of the main geometric parameters of objects formed by conical section. Contents 1 Ellipse 2 Parabola 3 Hyperbola ... Wikipedia

An artificial satellite is an orbital maneuver, the purpose of which (in the general case) is to transfer the satellite into an orbit with a different inclination. There are two types of such a maneuver: Changing the inclination of the orbit to the equator. Produced by inclusion ... ... Wikipedia

A branch of celestial mechanics that studies the motion of artificial space bodies: artificial satellites, interplanetary stations, and others spaceships. The scope of tasks of astrodynamics includes the calculation of the orbits of spacecraft, the determination of parameters ... ... Wikipedia

The Oberth effect in astronautics is the effect that a rocket engine moving at high speed produces more usable energy than the same engine moving slowly. The Oberth effect is caused by the fact that when ... ... Wikipedia

Customer ... Wikipedia

And the equipotential surfaces of a system of two bodies Lagrange points, libration points (lat. librātiō rocking) or L points ... Wikipedia

Books

Things of the twentieth century in drawings and photographs. Forward into space! Discoveries and achievements. Set of 2 books, . "Forward, into space! Discoveries and achievements" Since ancient times, man has dreamed of breaking away from the earth and conquering the sky, and then space. More than a hundred years ago, inventors were already thinking about creating ...
Onward to space! Discoveries and achievements, Klimentov Vyacheslav Lvovich, Sigorskaya Yulia Alexandrovna. Since ancient times, man has dreamed of breaking away from the earth and conquering the sky, and then space. More than a hundred years ago, inventors were already thinking about creating spaceships, but the beginning of space ...

Impulses along the axis of motion affect the shape and orientation* of the orbit and do not change its inclination.

Gravity maneuver as a natural phenomenon was first discovered by astronomers of the past, who realized that significant changes in the orbits of comets, their period (and hence their orbital speed) occur under the gravitational influence of the planets. So, after the transition of short-period comets from the Kuiper belt to the inner part of the solar system, a significant transformation of their orbits occurs precisely under the gravitational influence of massive planets, when exchanging angular momentum with them, without any energy costs.

Samu idea to use gravity maneuver for spaceflight was developed by Michael Minovich in the 60s, when, as a student, he did an internship at JPL *. The idea was quickly picked up and implemented in many space missions. But at first glance, the possibility of significantly accelerating the movement of the apparatus without consuming energy seems strange and requires explanation.

One often hears about the "capture" of asteroids and comets by the field of planets. Strictly speaking, capture without loss of energy is impossible: if a body approaches a massive planet, the modulus of its velocity first increases as it approaches, and then decreases by the same amount in the process of its removal. But the body can still go into the orbit of the planet's satellite if it decelerates (for example, there is deceleration in upper layers atmosphere, if the approach is close enough; or if significant tidal energy dissipation occurs; or, finally, if the destruction of the body occurs within the Roche limit with different velocity vectors acquired by the fragments). At the stage of formation of the solar system, an important factor was also the deceleration of the body in a gas-dust nebula. As for spacecraft, only in the case of launching a satellite into orbit, braking is used in the upper layers of the atmosphere (aerobraking). In a "pure" gravitational maneuver, the rule of equality of the velocity modulus before and after approaching the planet is strictly preserved (which was prompted by intuition: what you came with, you left with). What is the gain?

The gain becomes obvious if we move from planetocentric coordinates to heliocentric coordinates.

The most advantageous maneuvers are those of the giant planets, and they noticeably shorten the duration of the flight. Maneuvers are also used Earth and Venus, but this significantly increases the duration of space travel. All data in the table refer to the passive maneuver. But in some cases, at the pericenter of the flyby hyperbola, the spacecraft, with the help of its propulsion system, is given a small jet impulse, which gives a significant additional gain.

In flight, the device often does not need acceleration, but deceleration. It is easy to choose such a rendezvous geometry when the spacecraft velocity in heliocentric coordinates drops. It depends on the position of the velocity vectors in the exchange of angular momentum. Simplifying the problem, we can say that the approach of the apparatus to the planet from the inner side of its orbit leads to the fact that the apparatus gives the planet part of its angular momentum and slows down; and vice versa, approach from the outer side of the orbit leads to an increase in the momentum and speed of the spacecraft. It is interesting that no accelerometers on board can register a change in the vehicle's speed during maneuvers - they constantly record the state of weightlessness.

Advantages of a gravity assist maneuver compared to a Hohmann flight to the giant planets are so large that the payload of the apparatus can be doubled. As already mentioned, the time to reach the goal during a gravitational maneuver for massive giant planets is reduced very significantly. The development of the principles of the maneuver showed that less massive bodies (Earth, Venus and, in special cases, even the Moon) can also be used. Only the mass in a sense is exchanged for the flight time, which makes researchers wait 2-3 extra years. However, the desire to cut costs on expensive space program makes you come to terms with such a loss of time. Now the choice of the flight path is made, as a rule, multi-purpose, covering several planets. In 1986, a gravitational maneuver near Venus made it possible to ensure the meetings of the Soviet VEGA-1 and VEGA-2 spacecraft with Halley's comet.